Functions Ques 31
Let $f(x)=\left(1+b^{2}\right) x^{2}+2 b x+1$ and let $m(b)$ be the minimum value of $f(x)$. As $b$ varies, the range of $m(b)$ is
(2001, 1M)
(a) $[0,1]$
(b) $[0, \frac{1}{2}]$
(c) $[\frac{1}{2}, 1]$
(d) $[0,1]$
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Answer:
Correct Answer: 31.(d)
Solution:
Formula:
- Given, $f(x)=\left(1+b^{2}\right) x^{2}+2 b x+1$
$ =\left(1+b^{2}\right)( x+{\frac{b}{1+b^{2}}})^{2}+1-\frac{b^{2}}{1+b^{2}} $
$m(b)=$ minimum value of $f(x)=\frac{1}{1+b^{2}}$ is positive
and $m(b)$ varies from 1 to 0 , so range $=[0,1]$
BETA
